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This question showed up on my Practice exam.

A spaceship of mass $M_0$ (ship+fuel) travels horizontally in the earth's atmosphere, which exerts a drag force $F=-bv$. The ship is fitted with a retro-rocket pointed in the Forward direction (on front of ship, the purpose is to slow down the ship). The retrorocket burns fuel at constant rate$-\frac {\delta m}{\delta t}=k$, the exhaust leaves the nozzle at velocity u with respect to the rocket velocity. The spaceship enters the atmosphere at velocity $V_0$. Solve the velocity with respect to time.

The correct answer is: $$v=(\frac{ku}{b}+V_0)(\frac{m_0-kt}{m_0})^{(\frac{b}{k})}-\frac{ku}{b}$$

My attempt:

$$P_{final} = m_{final}*v_{final}$$ $$P_{final}=m_{rocket}*v_{rocket}-m_{fuel}*v_{fuel}-\int_t drag$$ To be honest, I am not too sure what to do with the Drag Force. but I know that $F=\frac {\delta p}{\delta t}$ so in order to get all the terms into momentum. I decided to take the integral.

$$m_{rocket}= m_0-m_{fuel}=m_0-\frac {\delta m}{\delta t}=m_0+kt$$ $$v_{rocket}= V_0 -\delta v$$ There must be some decrease in velocity here(?) $$m_{fuel}= \frac {\delta m}{\delta t}*t= -kt$$ $$v_{fuel} = V_0-\delta v+u$$ $$m_{final} = m_0+kt$$ Putting it all together: $$(m_0+kt)*V_{final}=(m_0+kt)(V_0-\delta v)-(kt)(V_0-\delta v+u)-\int_t bv$$ And I'm already pretty lost. I don't know how to deal with the drag force, and also don't know how to calculate $\delta v$. My answer doesn't look anything like the given solution.

The work for the solution or any pointers would be much appreciated.

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Let the rocket be travelling with a velocity $v+dv$ and let its mass be $m+dm$ where $dv$ and $dm$ is the decrease in velociy of the rocket and the mass that is ejected respectively after time $dt$.

$\frac{dm}{dt} = -k\implies m = M_0 - kt$

The change in linear momentum of the rocket mass system is:

$dp = m*v+dm*(v+u) - (m+dm)*(v+dv)$
$ \implies dp = u*dm - m*dv $

The change in momentum is due to the impulse provided by the drag force which is $-bv*dt$

Thus: $-bv*dt = u*dm - m*dv$
$ \implies -bv = uk- m*\frac{dv}{dt}$
Substituting $m = M_0 - kt$,

$-bv = uk- (M_0 - kt)*\frac{dv}{dt} \implies \frac{dv}{uk+bv} = \frac{dt}{M_0 - kt} $
Integrating from $V_0$ to $v$ and $0$ to $t$, we get the required answer

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